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`Plane Trigonometry De Moivre's Theorem Expansion of cos𝑛𝜃, ⋯ in powers sin𝜃 and cos𝜃 Expansion of sine and cosine in powers the angle Expansion of cos𝑛𝜃 and sin𝑛𝜃 in cosines or sines of multiples of 𝜃 Expansion of cos𝑛𝜃 and sin𝑛𝜃 in powers of sin𝜃 Expansion of cos𝑛𝜃 in descending powers of cos𝜃 Expansion of 𝜃 in powers of tan𝜃 (Gregory's series) formula for the calculation of the value of 𝜋 by Gregor's series To Prove that 𝜋 is Incommensurable Expansion of the sine and cosine in factors De Moivre's Property of the Circle Cotes's Properties Sources and References`

# Plane Trigonometry

## De Moivre's Theorem

756 (cos𝛼+𝑖sin𝛼)(cos𝛽+𝑖cos𝛽)⋯=cos(𝛼+𝛽+𝛾+⋯)+𝑖sin(𝛼+𝛽+𝛾+⋯), where 𝑖=−1 Proved by Induction 757 (cos𝜃+𝑖sin𝜃)𝑛=cos𝑛𝜃+𝑖sin𝑛𝜃 Proof: By Induction, or by putting 𝛼, 𝛽, ⋯ each = 𝜃 in (756).

## Expansion of cos𝑛𝜃, ⋯ in powers sin𝜃 and cos𝜃

758 cos𝑛𝜃=cos𝑛𝜃−𝐶(𝑛,2)cos𝑛−2𝜃sin2𝜃+𝐶(𝑛,4)cos𝑛−4𝜃sin4𝜃−⋯ 759 sin𝑛𝜃=𝑛cos𝑛−1𝜃sin𝜃−𝐶(𝑛,3)cos𝑛−3𝜃sin3𝜃+⋯ Proof: Expand (757) by Bin. Th., and equate real and imaginary parts. 760 tan𝑛𝜃=𝑛tan𝜃−𝐶(𝑛,3)tan3𝜃+⋯1−𝐶(𝑛,2)tan2𝜃+𝐶(𝑛,4)tan4𝜃−⋯ In series (758, 759), stop at, and exclude, all terms with indices greater than 𝑛. Note, 𝑛 is here an integer. 761 Let 𝑠𝑟=sum of the 𝐶(𝑛,𝑟) products of tan𝛼, tan𝛽, tan𝛾, ⋯ to 𝑛 terms. sin(𝛼+𝛽+𝛾+⋯)=cos𝛼cos𝛽⋯(𝑠1−𝑠3+𝑠5−⋯) 762 cos(𝛼+𝛽+𝛾+⋯)=cos𝛼cos𝛽⋯(1−𝑠2+𝑠4−⋯) Proof: By equating real and imaginary parts in (756). 763 tan(𝛼+𝛽+𝛾+⋯)=(𝑠1−𝑠3+𝑠5−𝑠7+⋯)1−𝑠2+𝑠4−𝑠6+⋯

## Expansion of sine and cosine in powers the angle

764 sin𝜃=𝜃−𝜃33!+𝜃55!−⋯, cos𝜃=1−𝜃22!+𝜃44!−⋯ Proof: Put 𝜃𝑛 for 𝜃 in (757) and 𝑛=∞, employing (754) and (755). 766 𝑒𝑖𝜃=cos𝜃+𝑖sin𝜃, 𝑒−𝑖𝜃=cos𝜃−𝑖sin𝜃By 150 768 𝑒𝑖𝜃+𝑒−𝑖𝜃=2cos𝜃, 𝑒𝑖𝜃−𝑒−𝑖𝜃=2𝑖sin𝜃 770 𝑖tan𝜃=𝑒𝑖𝜃−𝑒−𝑖𝜃𝑒𝑖𝜃+𝑒−𝑖𝜃, 1+𝑖tan𝜃1−𝑖tan𝜃=𝑒2𝑖𝜃

## Expansion of cos𝑛𝜃 and sin𝑛𝜃 in cosines or sines of multiples of 𝜃

772 2𝑛−1cos𝑛𝜃=cos𝑛𝜃+𝑛cos(𝑛−2)𝜃+𝐶(𝑛,2)cos(𝑛−1)𝜃+𝐶(𝑛,3)cos(𝑛−6)𝜃+⋯ 773 When 𝑛 is even, 2𝑛−1(−1)12𝑛sin𝑛𝜃=cos𝑛𝜃−𝑛cos(𝑛−2)𝜃+𝐶(𝑛,2)cos(𝑛−4)𝜃−𝐶(𝑛,3)cos(𝑛−6)𝜃+⋯ 774 And when 𝑛 is odd, 2𝑛−1(−1)𝑛−12sin𝑛𝜃=sin𝑛𝜃−𝑛sin(𝑛−2)𝜃+𝐶(𝑛,2)sin(𝑛−4)𝜃−𝐶(𝑛,3)sin(𝑛−6)𝜃+⋯ Observe that in these series the coefficients are those of the Binomial Theorem, with this exception: If 𝑛 be even, the last term must be divided by 2.
The series are obtained by expanding (𝑒𝑖𝜃±𝑒−𝑖𝜃)𝑛 by the Binomial Theorem, collecting the equidistant terms in pairs, and employing (768) and (769).

## Expansion of cos𝑛𝜃 and sin𝑛𝜃 in powers of sin𝜃

775 When 𝑛 is even, cos𝑛𝜃=1−𝑛22!sin2𝜃+𝑛2(𝑛2−22)4!sin4𝜃−𝑛2(𝑛2−22)(𝑛2−42)6!sin6𝜃+⋯ 776 When 𝑛 is odd, cos𝑛𝜃=cos𝜃1−𝑛2−12!sin2𝜃+(𝑛2−1)(𝑛2−32)4!sin4𝜃−(𝑛2−1)(𝑛2−32)(𝑛2−52)6!sin6𝜃+⋯ 777 When 𝑛 is even, sin𝑛𝜃=cos𝜃sin𝜃−𝑛2−223!sin3𝜃+(𝑛2−22)(𝑛2−42)5!sin5𝜃−(𝑛2−22)(𝑛2−42)(𝑛2−62)7!sin7𝜃+⋯ 778 When 𝑛 is odd, sin𝑛𝜃=𝑛sin𝜃−𝑛(𝑛2−1)3!sin3𝜃+𝑛(𝑛2−1)(𝑛2−32)5!sin5𝜃−𝑛(𝑛2−1)(𝑛2−32)(𝑛2−52)7!sin7𝜃+⋯ Proof: By (758), we may assume, when 𝑛 is an even integer cos𝑛𝜃=1+𝐴2sin2𝜃+𝐴4sin4𝜃+⋯+𝐴2𝑟sin2𝑟𝜃+⋯ Put 𝜃+𝑥 for 𝜃, and in cos𝑛𝜃cos𝑛𝑥−sin𝑛𝜃sin𝑛𝑥 substitute for cos𝑛𝑥 and sin𝑛𝑥 their values in powers of 𝑛𝑥 from (764). Each term on the right is of the type 𝐴2𝑟(sin𝜃cos𝑥+cos𝜃sin𝑥)2𝑟. Make similar substitutions for cos𝑥 and sin𝑥 in powers of 𝑥. Collect the two coefficients of 𝑥2 in each term by the multinomial theorem (137) and equate them all to the coefficient of 𝑥2 on the left. In this equation write cos2𝜃 for 1−sin2𝜃 everywhere, and then equate the coefficients of sin2𝑟𝜃 to obtain the relation between the successive equatities 𝐴2𝑟 and 𝐴2𝑟+2 for the series (775).
When 𝑛 is an odd integer, begin by assuming, by (759) sin𝑛𝜃=𝐴1sin𝜃+𝐴3sin3𝜃+⋯ 779 The expansions of cos𝑛𝜃 and sin𝑛𝜃 in powers of cos𝜃 are obtained by changing 𝜃 into 12𝜋−𝜃 in (775) to (778).

## Expansion of cos𝑛𝜃 in descending powers of cos𝜃

780 2cos𝑛𝜃=(2cos𝜃)𝑛−𝑛(2cos𝜃)𝑛−2+𝑛(𝑛−3)2!(2cos𝜃)𝑛−4−⋯+(−1)𝑟𝑛(𝑛−r−1)(𝑛−r−2)⋯(𝑛−2r+1)r!(2cos𝜃)𝑛−2r+⋯ up to the last positive power of 2cos𝜃.
Proof: By expanding each term of the identity log(1−𝑥𝑧)+log1−𝑧𝑥=log1−𝑧𝑥+1𝑥−𝑧 By (156), equating coefficients of 𝑧𝑛, and substituting from (768). 783 sin𝛼+𝑐sin(𝛼+𝛽)+𝑐2sin(𝛼+2𝛽)+⋯ to 𝑛 terms =sin𝛼−𝑐sin(𝛼−𝛽)−𝑐𝑛sin(𝛼+𝑛𝛽)+𝑐𝑛+1sin{𝛼+(𝑛−1)𝛽}1−2𝑐cos𝛽+𝑐2 784 If 𝑐 be < 1 and 𝑛 infinite, this becomes =sin𝛼−𝑐sin(𝛼−𝛽)1−2𝑐cos𝛽+𝑐2 785 cos𝛼+𝑐cos(𝛼+𝛽)+𝑐2cos(𝛼+2𝛽)+⋯ to 𝑛 terms = a similar result, changing sin into cos in the numerator. 786 similarly when 𝑐 is < 1 and 𝑛 infinite. 787 Method of summation: Substitute for the sines or cosines their exponential values (768). Sum the two resulting geometrical series, and substitute the sines or cosines again for the exponential values by (766). 788 𝑐sin(𝛼+𝛽)+𝑐22!sin(𝛼+2𝛽)+𝑐33!sin(𝛼+3𝛽)+⋯ to infinity =𝑒𝑐cos𝛽sin(𝛼+𝑐sin𝛽)−sin𝛼 789 𝑐cos(𝛼+𝛽)+𝑐22!cos(𝛼+2𝛽)+𝑐33!cos(𝛼+3𝛽)+⋯ to infinity =𝑒𝑐cos𝛽cos(𝛼+𝑐sin𝛽)−cos𝛼 Obtained by the rule in (787) 790 If, in the series (783) to (789), 𝛽 be changed into 𝛽+𝜋, the signs of the alternate terms will thereby be changed.

## Expansion of 𝜃 in powers of tan𝜃 (Gregory's series)

791 𝜃=tan𝜃−tan3𝜃3+tan5𝜃5−⋯ The series converges if tan𝜃 be not >1. Proof: By expanding the logarithm of the value of 𝑒2𝑖𝜃 in (771) by (158).

## formula for the calculation of the value of 𝜋 by Gregor's series

792 𝜋4=tan−112+tan−113=tan−115tan−11239791 794 𝜋4=4tan−115tan−1170+tan−1199 Proof: By employing the formula for tan(𝐴±𝐵), (631)

## To Prove that 𝜋 is Incommensurable

795 Convert the value of tan𝜃 in terms of 𝜃 from (764) and (765) into a continued fraction, thus tan𝜃=𝜃1𝜃23𝜃25𝜃27⋯; or this result may be obtained by putting 𝑖𝜃 for 𝑦 in (294), and by (770). Hence 1−𝜃tan𝜃=𝜃23𝜃25𝜃27 Put 𝜋2 for , and assume that 𝜋, and therefore 𝜋24, is commensurable. Let 𝜋24=𝑚𝑛, 𝑚 and 𝑛 being integers. Then we shall have 1=𝑚2𝑛𝑚𝑛5𝑛𝑚𝑛7𝑛
The continued fraction is incommensurable, by (177). But unity cannot be equal to an incommensurable quantity. Therefore 𝜋 is not commensurable. 796 If sin𝑥=𝑛sin(𝑥+𝛼), 𝑥=𝑛sin𝛼+𝑛22sin2𝛼+𝑛33sin3𝛼+⋯ 797 If tan𝑥=𝑛tan𝑦, 𝑥=𝑦−𝑚sin2𝑦+𝑚22sin4𝑦−𝑚33sin6𝑦+⋯, where 𝑚=1−𝑛1+𝑛
Proof:By substiuting the exponential values of the sine or tangent (769) and (770), and then eliminating 𝑥. 798 Coefficient of 𝑥𝑛 in the expansion of 𝑒𝑎𝑥cos𝑏𝑥=(𝑎2+𝑏2)𝑛2𝑛!cos𝑛𝜃, where 𝑎=𝑟cos𝜃 and 𝑏=𝑟sin𝜃.
For proof, substitute for cos𝑏𝑥 from (768); expand by (150); put 𝑎=𝑟cos𝜃 and 𝑏=𝑟sin𝜃 in the coefficient of 𝑒𝑥, employ (757). 799 When 𝑒<1, 1−𝑒21−𝑒cos𝜃=1+2𝑏cos𝜃+2𝑏2cos2𝜃+2𝑏3cos3𝜃+⋯, where 𝑏=𝑒1+1−𝑒2
For proof, put 𝑒=2𝑏1+𝑏2 and 2cos𝜃=𝑥+1𝑥, expand the fraction in two series of powers of 𝑥 by the method of (257), and substitute from (768). 800 sin𝛼+sin(𝛼+𝛽)+sin(𝛼+2𝛽)+⋯+sin{𝛼+(𝑛−1)𝛽}=sin𝛼+𝑛−12𝛽sin𝑛2𝛽sin𝛽2 801 cos𝛼+cos(𝛼+𝛽)+sin(𝛼+2𝛽)+⋯+cos{𝛼+(𝑛−1)𝛽}=cos𝛼+𝑛−12𝛽sin𝑛2𝛽sin𝛽2 802 If the terms in these series have the signs + and − alternately, change 𝛽 into 𝛽+𝜋 in the results.
Proof: Multiply the series by 2sin𝛽2, and apply (669) and (666). 803 If 𝛽=2𝜋𝑛 in (800) and (801), each series vanishes. 804 Generally, if 𝛽=2𝜋𝑛, and if 𝑟 be an integer not a multiple of 𝑛, the sum of the 𝑟th powers of the sines or cosines in (800) or (801) is zero if 𝑟 be odd; and if 𝑟 be even it is =𝑛2𝑟; by (772) to (774) 805 General Theorem: Denoting the sum of the series 𝑐+𝑐1𝑥+𝑐2𝑥2+⋯+𝑐𝑛𝑥𝑛 by 𝐹(𝑥); then 𝑐cos𝛼+𝑐1cos(𝛼+𝛽)+⋯+𝑐𝑛cos(𝛼+𝑛𝛽)=12{𝑒𝑖𝛼𝐹(𝑒𝑖𝛽)+𝑒−𝑖𝛼𝐹(𝑒−𝑖𝛽)} and 806 𝑐sin𝛼+𝑐1sin(𝛼+𝛽)+⋯+𝑐𝑛sin(𝛼+𝑛𝛽)=12𝑖{𝑒𝑖𝛼𝐹(𝑒𝑖𝛽)−𝑒−𝑖𝛼𝐹(𝑒−𝑖𝛽)} Provd by substituting for the sines and cosines their exponential values (766), ⋯.

## Expansion of the sine and cosine in factors

807 𝑥2𝑛−2𝑥𝑛𝑦𝑛cos𝑛𝜃+𝑦2𝑛=𝑥2−2𝑥𝑦cos𝜃+𝑦2𝑥2−2𝑥𝑦cos𝜃+2𝜋𝑛+𝑦2 to 𝑛 factors, adding 2𝜋𝑛 to the angle successively.
Proof: By solving the quadratic on the left, we get 𝑥=𝑦(cos𝑛𝜃+𝑖sin𝑛𝜃)1𝑛. The 𝑛 values of 𝑥 are found by (757) and (626), and thence tha factors. For the factors 𝑥𝑛±𝑦𝑛 see (480). 808 sin𝑛𝜙=2𝑛−1sin𝜙sin𝜙+𝜋𝑛sin𝜙+2𝜋𝑛 as far as 𝑛 factors of sines.
Proof: By putting 𝑥=𝑦=1 and 𝜃=2𝜙 in the last. 809 If 𝑛 be even, sin𝑛𝜙=2𝑛−1sin𝜙cos𝜙sin2𝜋𝑛sin2𝜙sin22𝜋𝑛sin2𝜙 810 If 𝑛 be odd, omit cos𝜙 and make up 𝑛 factors, reckoning two factors for each pair of terms in brackets.
Proof: From (808), by collecting equidistant factors in pairs, and applying (659). 811 cos𝑛𝜙=2𝑛−1sin𝜙+𝜋2𝑛sin𝜙+3𝜋2𝑛⋯ to 𝑛 factors. Proof: Put 𝜙+𝜋2𝑛 for 𝜙 in (808). 812 Also, if 𝑛 be odd, cos𝑛𝜙=2𝑛−1cos𝜙sin2𝜋2𝑛sin2𝜙sin23𝜋2𝑛sin2𝜙 813 If 𝑛 be even, omit cos𝜙,
Proof: as in (809) 814 𝑛=2𝑛−1sin𝜋𝑛sin2𝜋𝑛sin3𝜋𝑛sin(𝑛−1)𝜋𝑛 Proof: divide (809) by sin𝜙, and make 𝜙 vanish; then apply (754). 815 sin𝜃=𝜃1−𝜃𝜋21−𝜃2𝜋21−𝜃3𝜋2 816 cos𝜃=1−2𝜃𝜋21−2𝜃3𝜋21−2𝜃5𝜋2 Proof: Put 𝜙=𝜃𝑛 in (809) and (842); divide by (814) and make 𝑛 infinite. 817 𝑒𝑥−2cos𝜃+𝑒−𝑥=4sin2𝜃21+𝑥2𝜃21+𝑥2(2𝜋±𝜃)21+𝑥2(4𝜋±𝜃)2 Proved by substituting 𝑥=1+𝑧2𝑛, 𝑦=1−𝑧2𝑛, and 𝜃𝑛 for 𝜃 in (807) Making 𝑛 infinite, and reducing one series of factors to 4sin2𝜃2 by putting 𝑧=0.

## De Moivre's Property of the Circle

Circle: Take 𝑃 any point, and 𝑃𝑂𝐵=𝜃 any angle, 𝐵𝑂𝐶=𝐶𝑂𝐷=⋯=2𝜋𝑛; 𝑂𝑃=𝑥; 𝑂𝐵=𝑟 819 𝑥2𝑛−2𝑥𝑛𝑟𝑛cos𝑛𝜃+𝑟2𝑛=𝑃𝐵2𝑃𝐶2𝑃𝐷2⋯ to 𝑛 factors By (807) and (702), since 𝑃𝐵2=𝑥2−2𝑥𝑟cos𝜃+𝑟2, ⋯ 820 If 𝑥=𝑟, 2𝑟𝑛sin𝑛𝜃2=𝑃𝐵⋅𝑃𝐶⋅𝑃𝐷⋯ 821

## Cotes's Properties

If 𝜃=2𝜋2, 𝑥𝑛∼𝑟𝑛=𝑃𝐵⋅𝑃𝐶⋅𝑃𝐷⋯ 822 𝑥𝑛+𝑟𝑛=𝑃𝑎⋅𝑃𝑏⋅𝑃𝑐⋯

## Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive

ID: 210900012 Last Updated: 9/12/2021 Revision: 0 Ref: References

1. B. Joseph, 1978, University Mathematics: A Textbook for Students of Science &amp; Engineering
2. Ayres, F. JR, Moyer, R.E., 1999, Schaum's Outlines: Trigonometry
3. Hopkings, W., 1833, Elements of Trigonometry  Home 5

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